# For x>1,

Question:

For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x}$ is equal to :

1. (1) $\frac{x \log _{e} 2 x-\log _{e} 2}{x}$

2. (2) $\log _{\mathrm{e}} 2 \mathrm{x}$

3. (3) $\frac{x \log _{e} 2 x+\log _{e} 2}{x}$

4. (4) $x \log _{e} 2 x$

Correct Option: 1

Solution:

Consider the equation,

$(2 x)^{2 y}=4 e^{2 x-2 y}$

Taking log on both sides

$2 y \ln (2 x)=\ln 4+(2 x-2 y)$ $\ldots(1)$

Differentiating both sides w.r.t. $x$,

$2 y \frac{1}{2 x} 2+2 \ln (2 x) \frac{d y}{d x}=0+2-2 \frac{d y}{d x}$

$2 \frac{d y}{d x}\left(1+\ln (2 x)=2-\frac{2 y}{x}=\frac{2 x-2 y}{x}\right.$ ....(2)

From (1) and (2),

$\frac{d y}{d x}(1+\ln 2 x)=1-\frac{1}{x}\left(\frac{\ln 2+x}{1+\ln 2 x}\right)$

$\Rightarrow \quad(1+\ln 2 x)^{2} \frac{d y}{d x}=1+\ln (2 x)-\left(\frac{x+\ln 2}{x}\right)$

$=\frac{x \ln (2 x)-\ln 2}{x}$